本题题解 题目传送门:https://www.luogu.org/problem/P2303 给定一个整数\(n\),求 \[ \sum_{i=1}^n \gcd(n,i) \] 蒟蒻随便yy了一下搞出来个\(O(\sqrt{n})\)的算法 这题数据怎么这么水 首先看到gcd我们就下意识的对它反演一波对吧 第一步 \[ \sum_{i=1}^n \gcd(n,i) = \sum_{d|n} \varphi(d) \frac{n}{d} \] 这里提供两种化法,得到的结果都是这个. 法一 根据欧…

BZOJ2705 SDOI2012 Longge的问题 Description Longge的数学成绩非常好,并且他非常乐于挑战高难度的数学问题.现在问题来了:给定一个整数N,你需要求出∑gcd(i, N)(1<=i <=N). Input 一个整数,为N. Output 一个整数,为所求的答案. Sample Input 6 Sample Output 15 HINT [数据范围] 对于60%的数据,0<N<=2160<N<=216. 对于100%的数据,0<N…

[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=2705 [题目大意] 求出∑gcd(i,N)(1<=i<=N) [题解] $∑_{i=1}^{N}gcd(i,N)$ $=∑_{i=1}^{N}∑_{d|gcd(i,N)}\phi(d)$ $=∑ \phi(d)∑ _{1=<i<=N \land d|i \land d|N}1$ $=∑_{d|N}\phi(d)\lfloor\frac{i}{d}\rfloor$ [代码…

题目链接 题意:求\(\sum_{i=1}^{n}\gcd(i,n)\) 首先可以肯定,\(\gcd(i,n)|n\). 所以设\(t(x)\)表示\(gcd(i,n)=x\)的\(i\)的个数. 那么答案很显然就是\(\sum_{d|n}t(d)*d\). 那么\(t(x)\)怎么求呢. \[t(x)=\sum_{i=1}^{n}[\gcd(i,n)=x]\] 因为若\(\gcd(x,y)=1\),则有\(\gcd(xk,yk)=k\). 所以 \[t(x)=\sum_{i=1}^{n}[\g…

题意 求$ \sum_{i=1}^n gcd(i,n) $ 给定 $n(1\le n\le 2^{32}) $. 链接 题解 欧拉函数 $φ(x)$ :1到x-1有几个和x互质的数. gcd(i,n)必定是n的一个约数. 若p是n的约数,那么gcd(i,n)==p的有$φ(n/p)$个数,因为要使gcd(i,n)==p,i/p和n/p必须是互质的. 那么就是求i/p和n/p互质的i在[1,n]里有几个,就等价于 1/p,2/p,...,n/p 里面有几个和n/p互质,即φ(n/p). 求和的话,…

Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. …

Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6918   Accepted: 2234 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now…

Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N.…

题目链接:传送门 题目需求: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. 这题就是上一篇博客的变形. 题目解析:首先先求出与N互质的个数,即N的欧拉函数值,之后分解出N的因子来,求解方法如下. 证明: 要求有多少个 i 满足gcd(i, N) = d 如果gcd(i, N) = d,则gcd(i/d, N/d) = 1 由于i <= N,所以 i/d <= N/d,…

Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9190   Accepted: 3073 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now…